(d)

We have,

$\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ}$

$=\left(\tan 81^{\circ}+\tan 9^{\circ}\right)-\left(\tan 63^{\circ}+\tan 27^{\circ}\right)$

$=\left(\cot 9^{\circ}+\tan 9^{\circ}\right)-\left(\cot 27^{\circ}+\tan 27^{\circ}\right)$

$=\left(\frac{\cos 9^{\circ}}{\sin 9^{\circ}}+\frac{\sin 9^{\circ}}{\cos 9^{\circ}}\right)-\left(\frac{\cos 27^{\circ}}{\sin 27^{\circ}}+\frac{\sin 27^{\circ}}{\cos 27^{\circ}}\right)$

$=\left(\frac{\cos ^2 9^{\circ}+\sin ^2 9^{\circ}}{\sin 9^{\circ} \cos 9^{\circ}}\right)-\left(\frac{\cos ^2 27^{\circ}+\sin ^2 27^{\circ}}{\sin 27^{\circ} \cos 27^{\circ}}\right)$

$=\left(\frac{2}{2 \sin 9^{\circ} \cos 9^{\circ}}\right)-\left(\frac{2}{2 \sin 27^{\circ} \cos 27^{\circ}}\right)$

$=\left[\frac{2}{\sin 18^{\circ}}-\frac{2}{\sin 54^{\circ}}\right]$

$=2\left[\frac{1}{\sin 18^{\circ}}-\frac{1}{\cos 36^{\circ}}\right]=2\left[\frac{4}{\sqrt{5}-1}-\frac{4}{\sqrt{5}+1}\right]$

$\left[\because \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}, \cos 36^{\circ}=\frac{\sqrt{5}+1}{4}\right]$

$=8\left[\frac{\sqrt{5}+1-\sqrt{5}+1]}{5-1}\right]=4$